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I'm trying to find how to express the following:
I think it would be something like: $$(A - (B \cup C)) \cup (B - (A \cup C)) \cup (C - (A \cup B))$$ But does this have a name? A simpler way to express it? Since this expression grows fast while one add more sets. |
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I don't know of a name for this, but one way to express it would be $$\bigcup\limits_i A_i \setminus \bigcup\limits_{i\ne j} \left(A_i \cap A_j\right)$$ You don't need to specify more complicated intersections, because they're already included in what's being subtracted here. This captures the idea that it's intersections that you don't want. This also provides for a succinct description of it in words: "the union minus the pairwise intersections". |
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This is similar to, but distinct from the symmetric difference. If your collection is $\mathcal{C}=\{A,B,C,\ldots\}$, then you could write this set with cumbersome notation along the lines of $\left\{a\in\bigcup \mathcal{C}\left|\,\left|\left\{A\in\mathcal{C}\mid a\in A\right\}\right|=1\right.\right\}$. It would probably be clearer to use words, like "$\nabla \mathcal{C}$ will denote the set of all elements that are in exactly one set in the collection $\mathcal{C}$. |
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Since the OP's picture didn't contain a square 'holding' the circles, we pedantically construct our own universal set $U$, $\tag 1 U = \bigcup\limits_{i=1}^n A_i$ and let $\chi_{A_i}$ be the corresponding characteristic (indicator) functions. Set $\tag 2 C = \sum_{i=1}^n \chi_{A_i}$. It is easy to see that the the simple function $C$ can only take on values contained in $\{1,2,\cdots,n\}$, so $\tag 3 C:U \to \{1,2,\cdots,n\}$
Observe that in general, this inverse image can be the empty set. Consider for example these three sets, $\qquad A_1 = \{1,2\}$, $A_2 = \{2,3\}$ and $A_3 = \{3,1\}$. Note: If you choose a larger universal set containing all the $A_i$, then $C(x)$ can take on the value $0$. That would happen whenever $x \notin \bigcup A_i$. Practically, it would be best when looking at any simple function expressed as in (2) to write, $\tag 4 C:U \to \{0,1,2,\cdots,n\}$ That is the common methodology. |
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I don't know of a specific name for this type of expression, but a simpler way to write it would be to use countable unions: Instead of labeling your three (or in larger cases, $n$) sets by $A$, $B$, and $C$, label them as $A_1$, $A_2$, and $A_3$. Then for $n$ sets, you could write: $$\bigcup\limits_{i=1}^n \left(A_i - \bigcup\limits_{1 \leq i \leq n; \, j \neq i} A_j\right)$$ or, alternatively: $$\bigcup\limits_{i=1}^n\bigcap\limits_{1 \leq i \leq n; \, j \neq i} \left(A_i - A_j\right)$$ |
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Can be written as $(A\bigtriangleup B\bigtriangleup C)-(A \cap B\cap C)$ where $\bigtriangleup$ is the symmetric difference operator. |
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"The set of elements unique to each set." |
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